The cellular chain group C_n^{CW}(X) for a CW complex X is the free abelian group on the n-cells. For CP^2, which has one 0-cell, one 2-cell, and one 4-cell, what is C_2^{CW}(CP^2)?
A0
BZ
CZ^2
DZ/2Z
CP^2 has exactly one 2-cell, so C_2^{CW}(CP^2) = Z, generated by that single 2-cell. Similarly C_0 = Z (one 0-cell), C_4 = Z (one 4-cell), and C_1 = C_3 = 0 (no 1-cells or 3-cells). Since the boundary map d_n: C_n → C_{n-1} can only go between adjacent groups, and all odd-dimensional groups are zero, every boundary map is zero. Therefore H_n^{CW}(CP^2) = C_n^{CW}(CP^2), giving H_0 = H_2 = H_4 = Z and all others zero.
Question 2 True / False
The cellular boundary map d_n: C_n^{CW} → C_{n-1}^{CW} is determined by the degrees of the attaching maps of the n-cells.
TTrue
FFalse
Answer: True
The coefficient of the (n-1)-cell e^{n-1}_β in d_n(e^n_α) is the degree of the composite map: S^{n-1} →^{φ_α} X^{n-1} → X^{n-1}/X^{n-2} ≅ ∨S^{n-1} →^{collapse} S^{n-1}_β. This composite collapses all (n-1)-cells except e^{n-1}_β to a point, and the resulting map S^{n-1} → S^{n-1} has a well-defined degree. This degree is the entry in the boundary matrix. Computing cellular homology reduces to computing degrees of maps between spheres — a tractable algebraic problem.
Question 3 Short Answer
Compute the cellular homology of the real projective plane RP^2, given the CW structure with one 0-cell, one 1-cell, and one 2-cell where the attaching map of the 2-cell wraps around the 1-cell twice.
Think about your answer, then reveal below.
Model answer: The cellular chain complex is 0 → Z →^{d_2} Z →^{d_1} Z → 0. The attaching map of the 2-cell wraps around the single 1-cell twice, so d_2 has degree 2: d_2(e^2) = 2·e^1. The attaching map of the 1-cell connects the 0-cell to itself, and d_1 = 0 (both endpoints are the same vertex, so d_1(e^1) = v - v = 0). Therefore: H_0 = Z/im(d_1) = Z/0 = Z, H_1 = ker(d_1)/im(d_2) = Z/2Z, H_2 = ker(d_2)/0 = 0. So H_*(RP^2; Z) = (Z, Z/2Z, 0, ...).
The degree-2 attaching map is the key: it produces the Z/2Z torsion in H_1 (the core loop of RP^2 has order 2 in homology, because traversing it twice bounds the 2-cell). The cellular computation is much faster than a simplicial or singular computation would be — three chain groups and two boundary maps versus the much larger simplicial chain complex.
Question 4 True / False
Cellular homology agrees with singular homology: H_n^{CW}(X) ≅ H_n(X) for every CW complex X.
TTrue
FFalse
Answer: True
This is a theorem proved using the long exact sequence of the pair (X^n, X^{n-1}) and excision. The key steps: H_n(X^n, X^{n-1}) ≅ Z^{number of n-cells} (by excision, collapsing X^{n-2} and identifying the relative homology with the homology of a wedge of spheres). The cellular boundary map is identified with the connecting homomorphism in the long exact sequence. The resulting cellular chain complex has the same homology as the singular chain complex, proved by induction on the skeleta using the five lemma.